XOR

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4619 Accepted Submission(s): 1612

Problem Description

XOR is a kind of bit operator, we define that as follow: for two binary base number A and B, let C=A XOR B, then for each bit of C, we can get its value by check the digit of corresponding position in A and B. And for each digit, 1 XOR 1 = 0, 1 XOR 0 = 1, 0 XOR 1 = 1, 0 XOR 0 = 0. And we simply write this operator as ^, like 3 ^ 1 = 2,4 ^ 3 = 7. XOR is an amazing operator and this is a question about XOR. We can choose several numbers and do XOR operatorion to them one by one, then we get another number. For example, if we choose 2,3 and 4, we can get 2^3^4=5. Now, you are given N numbers, and you can choose some of them(even a single number) to do XOR on them, and you can get many different numbers. Now I want you tell me which number is the K-th smallest number among them.

Input

First line of the input is a single integer T(T<=30), indicates there are T test cases.
For each test case, the first line is an integer N(1<=N<=10000), the number of numbers below. The second line contains N integers (each number is between 1 and 10^18). The third line is a number Q(1<=Q<=10000), the number of queries. The fourth line contains Q numbers(each number is between 1 and 10^18) K1,K2,......KQ.

Output

For each test case,first output Case #C: in a single line,C means the number of the test case which is from 1 to T. Then for each query, you should output a single line contains the Ki-th smallest number in them, if there are less than Ki different numbers, output -1.

思路：
求出一组线性基，第k大就是k的二进制位上为1的
位对应的所有基底异或起来。特判可以形成0的情况。

代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn=1e4+10;
vector<ll>G;
ll a[maxn],b[64];
int n,Q,zero;
void work()
{
    zero=0;
    memset(b,0,sizeof(b));
    for(int i=1;i<=n;i++)
    {
        for(int j=62;j>=0;j--)
        {
            if(a[i]>>j&1)
            {
                if(b[j]) a[i]^=b[j];
                else
                {
                    b[j]=a[i];
                    for(int k=j-1;k>=0;k--) if(b[k]&&(b[j]>>k&1)) b[j]^=b[k];
                    for(int k=j+1;k<=62;k++) if(b[k]>>j&1) b[k]^=b[j];
                    break;
                }
            }
        }
        if(a[i]) zero++;
    }
    if(zero!=n) zero=1;
    else zero=0;
    G.clear();
    for(int i=0;i<=62;i++) if(b[i]) G.push_back(b[i]);
}
int main()
{
    int T,cas=0;scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
        work();
        ll len=G.size();
        scanf("%d",&Q);
        printf("Case #%d:\n",++cas);
        while(Q--)
        {
            ll k;scanf("%lld",&k);
            k-=zero;
            ll ans=0;
            if(k>=(1LL<<len)) ans=-1;
            else
            {
                for(int i=0;i<len;i++) if(k>>i&1) ans^=G[i];
            }
            printf("%lld\n",ans);
        }
    }
    return 0;
}

hdu 3949 XOR(线性基)

XOR

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